PROPERTIES OF THE l'4'ljmerical FliNCTION Fs

Size: px
Start display at page:

Download "PROPERTIES OF THE l'4'ljmerical FliNCTION Fs"

Transcription

1 PROPERTIES OF THE l'4'ljmerical FliNCTION Fs by I. Blilceoiu, V. Seleacu, N. Virla Departamet of Mathematics, Uiversity of Craiova Craiova (1100), ROMANIA I this paper are studied some properties of the umerical fuctio Fs(x):N - {O,I} 4' N Fs(x) = L Sp(x), where 5 p (x) = S(p%) is the Smaradache O<pS.r p pme fuctio defied i [4]. Numerical example: Fs(5) = S(2 5 ) ~ 5(3 5 ) + S(5 5 ); Fs(6) = S(2 6 ) + 5(3 6 ) + 5(~). It is kow that: (p -1)r + 1 $ 5(pr) $ pr so(p -1)r < S(pr) $ pr. Tha (1) Where Jr( x) is the umber of prime umbers smaller or equal with x.. % 1 PROPOSITION 1: The sequece T( r) = 1 log Fs (x) + L --. has limit i=2 F:(l) Proof The iequality Fs(x) > r(p2+ 4- P:..r) - Jr(x» implies -logfs(x) < < -logr(a. + P P1I(%) - (r» < -logr((r)a. - ;r(r» = -logr -log 1Z'(x) -log(a. -1). Tha for x=i the iequality (1) become: < < =---- FS(i) i(pl +"'+P:(I) - Jr(i» i(pijr(i) - Jr(i» ijr(i)(pt -1).r 1 Tha T(x)<l-log(x)-logJr(x)-log(A-l)+L =2 m(i)(pt -1).r 1 A =2 =:- T(x) = l-logx-logjr(x)+ L =2 ljr(l) =:- lim T(x) $ 1 - lim logx-lim 10gJr(x) + lim i: -. _1_. =1-oo-oo+L=-oo. ;c~"" ;c... "" ;c... ~ ;c... oc 1=2 I Jr( I) PRoPOsmON 1. The equatio Fs(x) = Fs(x + 1) has o solutio for x en - to, I}. 6

2 Proof First we cosider that x-i is a prime umber with x > 2. I the particular case x = 2 we obtai Fs(2) = 5(22) = 4; Fs(3) = 5(2 3 ) + 5(3 3 ) = 4 T 9 = l3. So F 2 (2) < Fs(3). Next we shall write the iequalities: (2) U sig the reductio ad absurdum method we suppose that the equatio Fs (x) = Fs (x + 1) has solutio. From (2) results the iequalities From (3) results that: +7Z{x + 1) > 0. But Ptr(x+l) > 7Z{x + 1) so the diferece from above is egative for x> 0, ad we obtaied a cotradictio. So Fs ( x) = Fs ( x + 1) has o solutio for x + 1 a prime umber. Next, we demostrate that the equatio Fs ( x) = Fs (x + 1) has o solutio for x ad x + 1 both composite umbers. Let p be a prime umber satisfig coditios P > x ad P ~ x - I. Such P exists 2 accordig to Bertrad's postulate for every x E ~ - {O, I}. Tha i the factorial of the umber p( x-i), the umber p appears at least x times. So, we have S(pX) ~ p(x -1). But p(x-i) < px+p-x (if p> x) ad px+p-x=(p-i)(x+i)+i~s(px+l). 2 Therefore :3 p ~ x -1 so that S(pX) < S(px+l). Tha Fs(x) = S(p+ +S(pX)+ +S(p~X» Fs(x + 1) = S(ptl)+... +S(pX+l)+ +S(p~+;» > Fs(x) I coclusio Fs ( x + 1) > Fs (x) for x ad x + 1 composite umbers. If x is a prime umber 7Z{ x) = 7Z{ x + 1) ad the fact that the equatio Fs ( x) = Fs ( x + 1) has o solutio has the same demostratio as above. Fially the equatio Fs (x) = Fs (x + 1) has o solutio for ay x EN - {O, I}. PROPOSITION 3. The fuctio Fs(x) is strictly icreasig fuctio o its domai of defiitio. The proof of this property is justified by the propositio 2. PROPOSITION 4. Fs(x + y) > Fs(x) + Fs(Y) Vx,y EN - {O, I}. Proof Let X,Y EN - {O, I} ad we suppose x < y. Accordig to the defiitio of FS(x) we have: 7

3 F( ) S( x+y). S(,rTY)... S( X"' Y ). S(.r"'V) X + Y = PI T"'T P:r(x) ' P:r(.r)+!'... + P:ft~) T (4) 'S(.r+Y) S(.r+Y ) -r P1f(y) P1f(x+y) But from (1) we have the followig iequalities: A = (x + Y)(Pt P:r(.r) + P:r(x)+l P(x+y) - ;r(x + y» < F(x + y) $ ad X(PI P1f(x) - ;r(x» + Y(PI P:r(.r) +"'+P:r(X) +"'+P(y) - ;r(y» < F(x) + F(y) $ $ x(p,.+ + P1f(X» + yep, P,t(x) + P1l(x)+i ' ~.» = B (6) We proof that B < A. B < A <:::> x(p, P1l(r» + Y(A P1l(r» + Y(PJZ(r) PJZ(y») < x(a P:r(.r» + Y(A+"'~ P1f(.r» + x(p1l(r)+l P:r(r+y» - x(x + y) + + Y(P1l(x)+l P1l(y» + y( P.t(y)+l P1l(x+y» - y;r(x + y) <:::> x(p1l(r)+l+"'+p:r(x+y) - ;r(x+ y»+ Y(p,t(Y)+i+... P1E{r+y) - ;r(x+ y» > But P:r(r+y) ~ ;r(x + y) so that the iequality from above is true. CONSEQUENCE: FS(XY) > Fs(x) + Fs(Y) 'if x,y en -'{O,l} Because x ady en - {O,l} ad xy > x + Y tha Fs(XY) > Fs(x + y) > Fs(x) + Fs(Y) PROPOsmON 5. We try to fid lim FS() -+oo a We have Fs() = L S(pr') ad: O<p;S P!=p1fll! A + p,,+"'+p:() - ;r() a - I PI + P2+"'+P1r(II) a - 1 If a= 1 tha lim 1 - a (A+"'+p1l(II)-;r(» = lim (Pt+"'+p1f(II)-;r(» = +co ~ lim Fs(~) =+00 -+OO 11-+"" _ ~ 8

4 We cosider ow a > 1. We try to fid lim --"I=::.JI~ a - I ;'!() L P, - ;r() L P,!!( ) Let a = L P, - ;r() ad bl1 = a - 1 1=1 ad lim ~ applig Stolz - Cesaro: -+oc a - I :,(+l) L PI-;r(+l)- L PI+;r() 1=1 1=1 = ( + l)a-l _ a - I if ( + 1) is a prime 0, otherwise Let c = L Pi ad d = a - I. 1=1 Tha!!(+l) L PI-LPI C+1 - c = 1=1 1=1 = P!!(+I) = d + 1 -d ( + l)a-i _ a- I ( + l)a-i - a - I +1 if ( + l)a-1 _ a - I ( + I) is a prime 0, otherwise First we cosider the limit of the fuctio. lim x I 1 Co r-+co (x + It-I - x a - I = 1m = 0 lor a - 2 > 1 HCO ( a-l)[ (x + l)a-2 - x a - 2 ] We used the l'hospital theorem: I the same way we have lim x + 1 = 0 for a > 3. r-+co (x + l)a-i - x a - I So, for a > 3 we have:. PI+P2+"'+P-;r() ad hm a-i = 0 r-+co r PI + P P - 0 1m a-i -. r-+co So lim F() = o. r-+co a Fially lim F (:) = { 0 r-+"" +00 for a> 3 for a ~ 1 9

5 BmuOGRAPHY II) M. Adrei, C. Dumitrescu., V. Seleacu, L. T U\escu, ~ t Zafir [2} P. Groas Some remarks o the Smaradache Fuctio, Smaradache Fuctio Joural. Vol. 4, NO.1 (1994) 1-5; A ote o S(p), Smaradache Fuctio Joural. V. 2-3, No.1 (1993) 33~ (3) M. Adrei. 1. BaIaceoiu. C Dumitrescu., E.lUdescu. N. Radescu. V. Seleacu [4) F. Smaradache A liear combiatio with Smaradache Fuctio to obtai the Idetity, Proceedigs of 26 m Aual Iraia Mathematic Coferece Sbahid Babarar Uiversity of Kerma Kerma - Ira March A Fuctio i the Number Theory, A. Uiv. Tui~ Ser.StMat. VotXVIII, fase. 1(1980) 9,

Bertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.

Bertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n. Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,,

More information

An introduction to the Smarandache Double factorial function

An introduction to the Smarandache Double factorial function A itroductio to the Smaradache Double factorial fuctio Felice Russo Via A. Iifate 7 67051 Avezzao (Aq) Italy I [1], [2] ad [3] the Smaradache Double factorial fuctio is defied as: Sdf() is the smallest

More information

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime. MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,

More information

Exponential Functions and Taylor Series

Exponential Functions and Taylor Series Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 207 Outlie Revistig the Expoetial Fuctio Taylor Series

More information

Review Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = =

Review Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = = Review Problems ICME ad MS&E Refresher Course September 9, 0 Warm-up problems. For the followig matrices A = 0 B = C = AB = 0 fid all powers A,A 3,(which is A times A),... ad B,B 3,... ad C,C 3,... Solutio:

More information

A 2nTH ORDER LINEAR DIFFERENCE EQUATION

A 2nTH ORDER LINEAR DIFFERENCE EQUATION A 2TH ORDER LINEAR DIFFERENCE EQUATION Doug Aderso Departmet of Mathematics ad Computer Sciece, Cocordia College Moorhead, MN 56562, USA ABSTRACT: We give a formulatio of geeralized zeros ad (, )-discojugacy

More information

Math 61CM - Solutions to homework 3

Math 61CM - Solutions to homework 3 Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig

More information

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences. MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13

More information

About the use of a result of Professor Alexandru Lupaş to obtain some properties in the theory of the number e 1

About the use of a result of Professor Alexandru Lupaş to obtain some properties in the theory of the number e 1 Geeral Mathematics Vol. 5, No. 2007), 75 80 About the use of a result of Professor Alexadru Lupaş to obtai some properties i the theory of the umber e Adrei Verescu Dedicated to Professor Alexadru Lupaş

More information

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

PUTNAM TRAINING INEQUALITIES

PUTNAM TRAINING INEQUALITIES PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca

More information

University of Manitoba, Mathletics 2009

University of Manitoba, Mathletics 2009 Uiversity of Maitoba, Mathletics 009 Sessio 5: Iequalities Facts ad defiitios AM-GM iequality: For a, a,, a 0, a + a + + a (a a a ) /, with equality iff all a i s are equal Cauchy s iequality: For reals

More information

MAT1026 Calculus II Basic Convergence Tests for Series

MAT1026 Calculus II Basic Convergence Tests for Series MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real

More information

SMARANDACHE FUNCTION JOURNAL

SMARANDACHE FUNCTION JOURNAL Vol. 6, No. 1-2-3, 1995 ISSN 1053-4792 SMARANDACHE FUNCTION JOURNAL S(n) is the smallest integer such that S(n)! is divisible by n S(n) is the smallest integer such that S(n)! is divisible by n S(n) is

More information

Exponential Functions and Taylor Series

Exponential Functions and Taylor Series MATH 4530: Aalysis Oe Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 2017 MATH 4530: Aalysis Oe Outlie

More information

Different kinds of Mathematical Induction

Different kinds of Mathematical Induction Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}

More information

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled 1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how

More information

, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx)

, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx) Cosider the differetial equatio y '' k y 0 has particular solutios y1 si( kx) ad y cos( kx) I geeral, ay liear combiatio of y1 ad y, cy 1 1 cy where c1, c is also a solutio to the equatio above The reaso

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 3 9/11/2013. Large deviations Theory. Cramér s Theorem

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 3 9/11/2013. Large deviations Theory. Cramér s Theorem MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/5.070J Fall 203 Lecture 3 9//203 Large deviatios Theory. Cramér s Theorem Cotet.. Cramér s Theorem. 2. Rate fuctio ad properties. 3. Chage of measure techique.

More information

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms. [ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006

MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006 MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the

More information

It is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.

It is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function. MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied

More information

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its Ifiite Series. Defiitios & covergece Defiitio... Let {a } be a sequece of real umbers. a) A expressio of the form a + a +... + a +... is called a ifiite series. b) The umber a is called as the th term

More information

PAijpam.eu ON DERIVATION OF RATIONAL SOLUTIONS OF BABBAGE S FUNCTIONAL EQUATION

PAijpam.eu ON DERIVATION OF RATIONAL SOLUTIONS OF BABBAGE S FUNCTIONAL EQUATION Iteratioal Joural of Pure ad Applied Mathematics Volume 94 No. 204, 9-20 ISSN: 3-8080 (prited versio); ISSN: 34-3395 (o-lie versio) url: http://www.ijpam.eu doi: http://dx.doi.org/0.2732/ijpam.v94i.2 PAijpam.eu

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

MAS111 Convergence and Continuity

MAS111 Convergence and Continuity MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece

More information

Solutions to HW Assignment 1

Solutions to HW Assignment 1 Solutios to HW: 1 Course: Theory of Probability II Page: 1 of 6 Uiversity of Texas at Austi Solutios to HW Assigmet 1 Problem 1.1. Let Ω, F, {F } 0, P) be a filtered probability space ad T a stoppig time.

More information

n=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n

n=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n Series. Defiitios ad first properties A series is a ifiite sum a + a + a +..., deoted i short by a. The sequece of partial sums of the series a is the sequece s ) defied by s = a k = a +... + a,. k= Defiitio

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

FUNCTIONAL EQUATIONS WITH PRIME ROOTS FROM ARITHMETIC EXPRESSIONS FOR. BARRY BRENT Efmhurst, New York 11373

FUNCTIONAL EQUATIONS WITH PRIME ROOTS FROM ARITHMETIC EXPRESSIONS FOR. BARRY BRENT Efmhurst, New York 11373 FUNCTIONAL EQUATIONS WITH PRIME ROOTS FROM ARITHMETIC EXPRESSIONS FOR BARRY BRENT Efmhurst, New York 373. I this article, a geeralized form of Euler f s law cocerig the sigma fuctio will be obtaied ad

More information

DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS

DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS VERNER E. HOGGATT, JR. Sa Jose State Uiversity, Sa Jose, Califoria 95192 ad CALVIN T. LONG Washigto State Uiversity, Pullma, Washigto 99163

More information

IJITE Vol.2 Issue-11, (November 2014) ISSN: Impact Factor

IJITE Vol.2 Issue-11, (November 2014) ISSN: Impact Factor IJITE Vol Issue-, (November 4) ISSN: 3-776 ATTRACTIVITY OF A HIGHER ORDER NONLINEAR DIFFERENCE EQUATION Guagfeg Liu School of Zhagjiagag Jiagsu Uiversit of Sciece ad Techolog, Zhagjiagag, Jiagsu 56,PR

More information

CSE 1400 Applied Discrete Mathematics Number Theory and Proofs

CSE 1400 Applied Discrete Mathematics Number Theory and Proofs CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of

More information

Q-BINOMIALS AND THE GREATEST COMMON DIVISOR. Keith R. Slavin 8474 SW Chevy Place, Beaverton, Oregon 97008, USA.

Q-BINOMIALS AND THE GREATEST COMMON DIVISOR. Keith R. Slavin 8474 SW Chevy Place, Beaverton, Oregon 97008, USA. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 2008, #A05 Q-BINOMIALS AND THE GREATEST COMMON DIVISOR Keith R. Slavi 8474 SW Chevy Place, Beaverto, Orego 97008, USA slavi@dsl-oly.et Received:

More information

PROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.

PROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1. Math 7 Sprig 06 PROBLEM SET 5 SOLUTIONS Notatios. Give a real umber x, we will defie sequeces (a k ), (x k ), (p k ), (q k ) as i lecture.. (a) (5 pts) Fid the simple cotiued fractio represetatios of 6

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

HOMEWORK #10 SOLUTIONS

HOMEWORK #10 SOLUTIONS Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous

More information

2 Banach spaces and Hilbert spaces

2 Banach spaces and Hilbert spaces 2 Baach spaces ad Hilbert spaces Tryig to do aalysis i the ratioal umbers is difficult for example cosider the set {x Q : x 2 2}. This set is o-empty ad bouded above but does ot have a least upper boud

More information

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4. 11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although

More information

This section is optional.

This section is optional. 4 Momet Geeratig Fuctios* This sectio is optioal. The momet geeratig fuctio g : R R of a radom variable X is defied as g(t) = E[e tx ]. Propositio 1. We have g () (0) = E[X ] for = 1, 2,... Proof. Therefore

More information

Ma 530 Introduction to Power Series

Ma 530 Introduction to Power Series Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power

More information

COMMON FIXED POINT THEOREMS VIA w-distance

COMMON FIXED POINT THEOREMS VIA w-distance Bulleti of Mathematical Aalysis ad Applicatios ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 3, Pages 182-189 COMMON FIXED POINT THEOREMS VIA w-distance (COMMUNICATED BY DENNY H. LEUNG) SUSHANTA

More information

Lecture 7: Properties of Random Samples

Lecture 7: Properties of Random Samples Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ

More information

MDIV. Multiple divisor functions

MDIV. Multiple divisor functions MDIV. Multiple divisor fuctios The fuctios τ k For k, defie τ k ( to be the umber of (ordered factorisatios of ito k factors, i other words, the umber of ordered k-tuples (j, j 2,..., j k with j j 2...

More information

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals

More information

Find quadratic function which pass through the following points (0,1),(1,1),(2, 3)... 11

Find quadratic function which pass through the following points (0,1),(1,1),(2, 3)... 11 Adrew Powuk - http://www.powuk.com- Math 49 (Numerical Aalysis) Iterpolatio... 4. Polyomial iterpolatio (system of equatio)... 4.. Lier iterpolatio... 5... Fid a lie which pass through (,) (,)... 8...

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SCORE. Exam 2. MA 114 Exam 2 Fall 2016 MA 4 Exam Fall 06 Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use

More information

1+x 1 + α+x. x = 2(α x2 ) 1+x

1+x 1 + α+x. x = 2(α x2 ) 1+x Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem

More information

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x

More information

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1 MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The

More information

Recurrence Relations

Recurrence Relations Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial(-)); } Let t be the umber of multiplicatios eeded to calculate factorial(). The

More information

NEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE

NEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE UPB Sci Bull, Series A, Vol 79, Iss, 207 ISSN 22-7027 NEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE Gabriel Bercu We itroduce two ew sequeces of Euler-Mascheroi type which have fast covergece

More information

Oscillation and Property B for Third Order Difference Equations with Advanced Arguments

Oscillation and Property B for Third Order Difference Equations with Advanced Arguments Iter atioal Joural of Pure ad Applied Mathematics Volume 3 No. 0 207, 352 360 ISSN: 3-8080 (prited versio); ISSN: 34-3395 (o-lie versio) url: http://www.ijpam.eu ijpam.eu Oscillatio ad Property B for Third

More information

CHAPTER 1 SEQUENCES AND INFINITE SERIES

CHAPTER 1 SEQUENCES AND INFINITE SERIES CHAPTER SEQUENCES AND INFINITE SERIES SEQUENCES AND INFINITE SERIES (0 meetigs) Sequeces ad limit of a sequece Mootoic ad bouded sequece Ifiite series of costat terms Ifiite series of positive terms Alteratig

More information

Introduction to Extreme Value Theory Laurens de Haan, ISM Japan, Erasmus University Rotterdam, NL University of Lisbon, PT

Introduction to Extreme Value Theory Laurens de Haan, ISM Japan, Erasmus University Rotterdam, NL University of Lisbon, PT Itroductio to Extreme Value Theory Laures de Haa, ISM Japa, 202 Itroductio to Extreme Value Theory Laures de Haa Erasmus Uiversity Rotterdam, NL Uiversity of Lisbo, PT Itroductio to Extreme Value Theory

More information

MATH 31B: MIDTERM 2 REVIEW

MATH 31B: MIDTERM 2 REVIEW MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate x (x ) (x 3).. Partial Fractios Solutio: The umerator has degree less tha the deomiator, so we ca use partial fractios. Write x (x ) (x 3) = A x + A (x ) +

More information

7.1 Convergence of sequences of random variables

7.1 Convergence of sequences of random variables Chapter 7 Limit Theorems Throughout this sectio we will assume a probability space (, F, P), i which is defied a ifiite sequece of radom variables (X ) ad a radom variable X. The fact that for every ifiite

More information

Solutions for Math 411 Assignment #2 1

Solutions for Math 411 Assignment #2 1 Solutios for Math 4 Assigmet #2 A2. For each of the followig fuctios f : C C, fid where f(z is complex differetiable ad where f(z is aalytic. You must justify your aswer. (a f(z = e x2 y 2 (cos(2xy + i

More information

The Simplex algorithm: Introductory example. The Simplex algorithm: Introductory example (2)

The Simplex algorithm: Introductory example. The Simplex algorithm: Introductory example (2) Discrete Mathematics for Bioiformatics WS 07/08, G. W. Klau, 23. Oktober 2007, 12:21 1 The Simplex algorithm: Itroductory example The followig itroductio to the Simplex algorithm is from the book Liear

More information

ON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS

ON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS ON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS NORBERT KAIBLINGER Abstract. Results of Lid o Lehmer s problem iclude the value of the Lehmer costat of the fiite cyclic group Z/Z, for 5 ad all odd. By complemetary

More information

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral,

More information

McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems

McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz fuctios. Let Lip K be the set of all fuctios cotiuous fuctios o [, 1] satisfyig a Lipschitz

More information

Please do NOT write in this box. Multiple Choice. Total

Please do NOT write in this box. Multiple Choice. Total Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should

More information

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5 Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You

More information

Math 104: Homework 2 solutions

Math 104: Homework 2 solutions Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does

More information

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory 1. Graph Theory Prove that there exist o simple plaar triagulatio T ad two distict adjacet vertices x, y V (T ) such that x ad y are the oly vertices of T of odd degree. Do ot use the Four-Color Theorem.

More information

Properties and Tests of Zeros of Polynomial Functions

Properties and Tests of Zeros of Polynomial Functions Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by

More information

IIT JAM Mathematical Statistics (MS) 2006 SECTION A

IIT JAM Mathematical Statistics (MS) 2006 SECTION A IIT JAM Mathematical Statistics (MS) 6 SECTION A. If a > for ad lim a / L >, the which of the followig series is ot coverget? (a) (b) (c) (d) (d) = = a = a = a a + / a lim a a / + = lim a / a / + = lim

More information

Real Variables II Homework Set #5

Real Variables II Homework Set #5 Real Variables II Homework Set #5 Name: Due Friday /0 by 4pm (at GOS-4) Istructios: () Attach this page to the frot of your homework assigmet you tur i (or write each problem before your solutio). () Please

More information

Inverse Matrix. A meaning that matrix B is an inverse of matrix A.

Inverse Matrix. A meaning that matrix B is an inverse of matrix A. Iverse Matrix Two square matrices A ad B of dimesios are called iverses to oe aother if the followig holds, AB BA I (11) The otio is dual but we ofte write 1 B A meaig that matrix B is a iverse of matrix

More information

PRELIM PROBLEM SOLUTIONS

PRELIM PROBLEM SOLUTIONS PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems

More information

Equations and Inequalities Involving v p (n!)

Equations and Inequalities Involving v p (n!) Equatios ad Iequalities Ivolvig v (!) Mehdi Hassai Deartmet of Mathematics Istitute for Advaced Studies i Basic Scieces Zaja, Ira mhassai@iasbs.ac.ir Abstract I this aer we study v (!), the greatest ower

More information

Math 312 Lecture Notes One Dimensional Maps

Math 312 Lecture Notes One Dimensional Maps Math 312 Lecture Notes Oe Dimesioal Maps Warre Weckesser Departmet of Mathematics Colgate Uiversity 21-23 February 25 A Example We begi with the simplest model of populatio growth. Suppose, for example,

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you

More information

Implicit function theorem

Implicit function theorem Jovo Jaric Implicit fuctio theorem The reader kows that the equatio of a curve i the x - plae ca be expressed F x, =., this does ot ecessaril represet a fuctio. Take, for example F x, = 2x x =. (1 either

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim

Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim Real Aalysis Fall 004 Take Home Test SOLUTIONS. Use the defiitio of a limit to show that (a) lim si = 0 (b) Proof. Let ε > 0 be give. Defie N >, where N is a positive iteger. The for ε > N, si 0 < si

More information

INEQUALITIES BJORN POONEN

INEQUALITIES BJORN POONEN INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad

More information

Infinite Series and Improper Integrals

Infinite Series and Improper Integrals 8 Special Fuctios Ifiite Series ad Improper Itegrals Ifiite series are importat i almost all areas of mathematics ad egieerig I additio to umerous other uses, they are used to defie certai fuctios ad to

More information

Solutions to Tutorial 3 (Week 4)

Solutions to Tutorial 3 (Week 4) The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial Week 4 MATH2962: Real ad Complex Aalysis Advaced Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/

More information

M A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O

M A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O M A T H 2 4 0 F A L L 2 0 1 4 HOMEWORK ASSIGNMENT #4 CORRECTION Algebra I 1 4 / 1 0 / 2 0 1 4 U N I V E R S I T Y O F T O R O N T O P r o f e s s o r : D r o r B a r - N a t a Correctio Homework Assigmet

More information

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6 Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is

More information

1 Counting and Stirling Numbers

1 Counting and Stirling Numbers 1 Coutig ad Stirlig Numbers Natural Numbers: We let N {0, 1, 2,...} deote the set of atural umbers. []: For N we let [] {1, 2,..., }. Sym: For a set X we let Sym(X) deote the set of bijectios from X to

More information

(2) F j+2 = F J+1 + F., F ^ = 0, F Q = 1 (j = -1, 0, 1, 2, ). Editorial notes This is not our standard Fibonacci sequence.

(2) F j+2 = F J+1 + F., F ^ = 0, F Q = 1 (j = -1, 0, 1, 2, ). Editorial notes This is not our standard Fibonacci sequence. ON THE LENGTH OF THE EUCLIDEAN ALGORITHM E. P. IV1ERKES ad DAVSD MEYERS Uiversity of Ciciati, Ciciati, Ohio Throughout this article let a ad b be itegers, a > b >. The Euclidea algorithm geerates fiite

More information

CALCULATION OF FIBONACCI VECTORS

CALCULATION OF FIBONACCI VECTORS CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

Lecture 15: Consequences of Continuity. Theorem Suppose a; b 2 R, a<b, and f :[a; b]! R. If f is continuous and s 2 R is

Lecture 15: Consequences of Continuity. Theorem Suppose a; b 2 R, a<b, and f :[a; b]! R. If f is continuous and s 2 R is Lecture 15: Cosequeces of Cotiuity 15.1 Itermediate Value Theorem The followig result is kow as the Itermediate Value Theorem. Theorem Suppose a; b 2 R, a

More information

~j=zhax~ 6=0, t,2... ), k=~

~j=zhax~ 6=0, t,2... ), k=~ Numerische Mathematik 6, 35 -- 39 (964) Exactess Coditios i Numerical Quadrature* By HERBERT S. WILl* The itegral b () I--f t(x)dx is usually computed umerically by meas of a formula of the type (2) I

More information

Enumerative & Asymptotic Combinatorics

Enumerative & Asymptotic Combinatorics C50 Eumerative & Asymptotic Combiatorics Stirlig ad Lagrage Sprig 2003 This sectio of the otes cotais proofs of Stirlig s formula ad the Lagrage Iversio Formula. Stirlig s formula Theorem 1 (Stirlig s

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios

More information

Math 128A: Homework 1 Solutions

Math 128A: Homework 1 Solutions Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that

More information

( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!

( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =! .8,.9: Taylor ad Maclauri Series.8. Although we were able to fid power series represetatios for a limited group of fuctios i the previous sectio, it is ot immediately obvious whether ay give fuctio has

More information

Section 7 Fundamentals of Sequences and Series

Section 7 Fundamentals of Sequences and Series ectio Fudametals of equeces ad eries. Defiitio ad examples of sequeces A sequece ca be thought of as a ifiite list of umbers. 0, -, -0, -, -0...,,,,,,. (iii),,,,... Defiitio: A sequece is a fuctio which

More information

It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.

It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial. Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable

More information

11.5 Alternating Series, Absolute and Conditional Convergence

11.5 Alternating Series, Absolute and Conditional Convergence .5.5 Alteratig Series, Absolute ad Coditioal Covergece We have see that the harmoic series diverges. It may come as a surprise the to lear that ) 2 + 3 4 + + )+ + = ) + coverges. To see this, let s be

More information

Math 113 Exam 4 Practice

Math 113 Exam 4 Practice Math Exam 4 Practice Exam 4 will cover.-.. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for

More information

ANOTHER GENERALIZED FIBONACCI SEQUENCE 1. INTRODUCTION

ANOTHER GENERALIZED FIBONACCI SEQUENCE 1. INTRODUCTION ANOTHER GENERALIZED FIBONACCI SEQUENCE MARCELLUS E. WADDILL A N D LOUIS SACKS Wake Forest College, Wisto Salem, N. C., ad Uiversity of ittsburgh, ittsburgh, a. 1. INTRODUCTION Recet issues of umerous periodicals

More information